I would like to know how to deal with:
$$\mathcal{F} g(t-a(t))=??$$
Cause I know that:
$$\mathcal{F} g(t-b)=e^{-i2\pi f b}G(f)$$
where $b$ is a real constant and $G$ represents the Fourier transform of $g$.
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There are easy answers if $a(t)=\alpha t+\beta$ and none others in the tables of Fourier transforms (Erdelyi et al, Oberhettinger, Gradstyn & Ryzhik). Remark that there are very few formulas of this kind, for particular functions $a$ like $\sqrt{\cdot}$, with the Laplace transform in Prudnikov et al. So the answer is probably : no formulas are known except for linear functions $a$. – Tom-Tom Jan 20 '16 at 23:11
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by writing $g$ as an inverse Fourier transform, you get terms of the form $\int_{-\infty}^\infty e^{2 i \pi ((f_0-f)t- f_0 a(t))} dt$ from which at least we can estimate the bandwidth of $g(t-a(t))$ on small windows of time – reuns Jan 20 '16 at 23:21